What mass of oxygen gas is required for combustion of a #12.0*g# mass of propane?

1 Answer
Apr 3, 2017

Answer:

We need (i) a stoichiometric equation:

Explanation:

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)#

Is it balanced? It's your problem not mine.

And (ii) we need equivalent quantities of propane and dioxygen.

#"Moles of propane"=(12.0*g)/(44.1*g*mol^-1)=0.272*mol#

#"Moles of dioxygen"=(5.01*g)/(32.00*g*mol^-1)=0.156*mol#

Clearly, there is insufficient oxygen for complete combustion: we would require #5xx0.272*mol# dioxygen gas.

So what would occur? Well, this is a case of incomplete combustion. I could invoke a stoichiometric equation of the sort:

#C_3H_8(g) + 4O_2(g) rarr 2CO_2(g) + C + 4H_2O(l)#

OR

#C_3H_8(g) + 7/2O_2(g) rarr CO_2(g) +CO(g)+ C(s) + 4H_2O(l)#

Which are both still deficient with respect to the oxidant. This question is NOT well-proposed; certainly it is unsuitable for an A-level student, and it would tax many first year university students.

You could assume COMPLETE combustion on the basis of the first equation, but the product mix would likely have #C# and #CO#, the products of incomplete combustion. I take it you could work out the stoichiometric excess of propane on the basis of complete combustion? If not, raise a flag, and someone will readdress the problem.